160. Intersection of Two Linked Lists #

Problem #

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

Problem Summary #

Find the intersection point of 2 linked lists.

Solution Approach #

The idea for this problem is actually similar to finding a cycle in a linked list.

If the two given linked lists have the same length, just scan from the head to the end. If they are not the same length, you need to first “splice them together” to make them the same length. Append B to the end of A, and append A to the end of B. This way the lengths of the two linked lists are both A + B. Then scan in order and compare whether the nodes of the two linked lists are the same.

Code #

package leetcode

import "fmt"

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func getIntersectionNode(headA, headB *ListNode) *ListNode {
	//boundary check
	if headA == nil || headB == nil {
		return nil
	}

	a := headA
	b := headB

	//if a & b have different len, then we will stop the loop after second iteration
	for a != b {
		//for the end of first iteration, we just reset the pointer to the head of another linkedlist
		if a == nil {
			a = headB
		} else {
			a = a.Next
		}

		if b == nil {
			b = headA
		} else {
			b = b.Next
		}
		fmt.Printf("a = %v b = %v\n", a, b)
	}
	return a
}